A1020 Tree Traversals （25 分）

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

 #include<cstdio>
 #include<queue>
 using namespace std;
 int post[],in[];
 struct node{
     int data;
     node *lchild;
     node *rchild;
 };
 node *create(int postL,int postR,int inL,int inR){
     if(postL>postR){
         return NULL;
     }
     node *root=new node;
     root->data=post[postR];
     int k;
     for(k=inL;k<=inR;k++){
         if(post[postR]==in[k]){
             break;
         }
     }
     int numleft=k-inL;
     root->lchild=create(postL,postL+numleft-,inL,k-);//这里经常出错
     root->rchild=create(postL+numleft,postR-,k+,inR);
     return root;
 }
 void LayerOrder(node*root,int k){
     if(root==NULL){
         return;
     }
     queue<node*> q;
     q.push(root);
     while(!q.empty()){
         node *now=q.front();
         q.pop();
         if(k!=){
             printf("%d ",now->data);
             k--;
         }else{
             printf("%d",now->data);
         }
         if(now->lchild!=NULL){
             q.push(now->lchild);
         }
         if(now->rchild!=NULL){
             q.push(now->rchild);
         }
     }
 }
 int main(){
     int n;
     scanf("%d",&n);
     for(int i=;i<n;i++){
         scanf("%d",&post[i]);
     }
     for(int i=;i<n;i++){
         scanf("%d",&in[i]);
     }
     node *p=create(,n-,,n-);
     LayerOrder(p,n);
     return ;
 } 

Mist Note: 解决本题，本小编的思路是，先利用中序和后序序列重建二叉树，然后再层序遍历这棵树。

很多算法还是基础的，主要是在递归重建那里经常出问题。左子树和右子树的递归。

root->lchild=create(postL,postL+numleft-1,inL,k-);

root->rchild=create(postL+numleft,postR-,k+,inR);

红色部分经常出错，切记，我们在写代码的时候，经常是利用数组，而数组是以0开始的，而numleft是个数，你加之后，肯定需要-1，才符合定义，不要混淆。

本题还需要注意输出格式，最后层序遍历的最后一个节点，是不需要空格的。